Question

graph each equation of the system. Then solve the system to find the points of intersection.

$$\begin{gathered} 4x^2+3y^2=4 \\ 2x^2-6y^2=-3 \end{gathered}$$

Short answer

The point of intersection is $\left(\sqrt{\frac{1}{2}},\sqrt{\frac{2}{3}}\right),\left(\sqrt{\frac{1}{2}},-\sqrt{\frac{2}{3}}\right),\left(-\sqrt{\frac{1}{2}},\sqrt{\frac{2}{3}}\right),\left(-\sqrt{\frac{1}{2}},-\sqrt{\frac{2}{3}}\right)$.

Step-by-step solution

Step 1. Given information

$$\begin{gathered} 4x^2+3y^2=4 \\ 2x^2-6y^2=-3 \end{gathered}$$

Step 2. Graphing each equation

$2x^2-6y^2=-3$

$2x-3y+4=0$

Step 3. Calculation of intersection point

$$\begin{gathered} Consider, \\ 4x^2+3y^2=4.........\left(1\right) \\ 2x^2-6y^2=-3.......\left(2\right) \\ Addingeq.2toeq.1\times 2 \\ weget \\ 10x^2=5 \\ x=\pm \sqrt{\frac{1}{2}} \\ substitutex=\pm \sqrt{\frac{1}{2}}ineq.2togetx \\ x=\sqrt{\frac{1}{2}} \\ y=\pm \sqrt{\frac{2}{3}} \\ x=-\sqrt{\frac{1}{2}} \\ y=\pm \sqrt{\frac{2}{3}} \end{gathered}$$